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%I #16 Mar 17 2017 22:53:48
%S 32,3042,375131,46137317,5674515856,697919312217,85838400887831,
%T 10557425389890242,1298477484555612931,159702173174950499517,
%U 19642068823034355828656,2415814763060050816424417
%N Maximum element in the continued fraction for F(5n+3)^5/F(5n+2)^5 where F=A000045 are Fibonacci numbers.
%D B. Cloitre, On rational sequences yielding continued fractions with unbounded coefficients, in preparation
%H G. C. Greubel, <a href="/A113500/a113500.txt">Table of n, a(n) for n = 0..475</a>
%F a(n) = 2*L(10*n+4) + L(10*n+5) + (-1)^n*7 - 1, where L(k) denotes the k-th Lucas number L(k) = F(k-1) + F(k+1), for n >= 0.
%F Empirical g.f.: (x^4-140*x^3-965*x^2+894*x-32) / ((x-1)*(x+1)*(x^2-123*x+1)). - _Colin Barker_, Jun 17 2013
%t Table[2*LucasL[10*n + 4] + LucasL[10*n + 5] + 7*(-1)^n - 1, {n,0,50}] (* _G. C. Greubel_, Mar 13 2017 *)
%o (PARI) a(n)=vecmax(contfrac(fibonacci(5*n+3)^5/fibonacci(5*n+2)^5))
%Y Cf. A000045.
%K nonn
%O 0,1
%A _Benoit Cloitre_, Jan 10 2006
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